Q.1.24 - An Introduction to Thermal Physics

Question

Calculate the total thermal energy in a gram of lead at room temperature, assuming that none of the degrees of freedom are "frozen out" (this happens to be a good assumption in this case). 

Step-by-Step Solution

Answer

The total thermal energy$\mathrm{U}=35.3\mathrm{J}$.

Step 1. Introduction.

Each atom in lead has six degrees of freedom in terms of vibration ($3$ kinetic and $3$potential). There are no transitional degrees of freedom, and there are no rotational degrees of freedom for a single atom because the atoms are fixed in their locations in the lattice.

Step 2. Explanation.

Given the molar mass of lead, the number of molecules in one gram of lead is,

Here$\mathrm{u}=1.66\times 10{}^{27} \mathrm{kg}$.

$\begin{array}{rcl}\mathrm{N}& =& \frac{\text{ mass }}{\text{ molar mass }}\\ & =& \frac{1\times {10}^3 \mathrm{kg}}{207.2\times 1.66\times {10}^{27} \mathrm{kg}}\\ & =& 2.91\times {10}^{21}\end{array}$

The thermal energy at $\mathrm{T}=293\mathrm{K}$

$\begin{array}{rcl}\mathrm{U}& =& \frac{\mathrm{f}}{2}\mathrm{NkT}\\ & =& \frac{6}{2}\times 2.91\times {10}^{21}\times 1.38\times {10}^{23}\times 293\\ & =& 35.3 \mathrm{J}\end{array}$